# Video: US-SAT05S4-Q20-659123709815

Consider the function π(π₯, π¦) = (πΎπ₯Β³)/π¦. If πΎ is a constant in the definition of the function π and π(π, π) = 3, what is the value of π(2π, 3π)?

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### Video Transcript

Consider the function π of π₯, π¦ equals πΎ times π₯ cubed over π¦. If πΎ is a constant in the definition of the function π and π of π, π equals three, what is the value of π of two π, three π?

At first, it might not seem like we have enough information to solve this problem. But letβs write down what we know: π of π₯, π¦ equals πΎ times π₯ cubed over π¦ and π of π, π equals three. We could also say that π of π, π equals πΎ times π cubed over π. And since weβre interested in π of two π, three π, we can say that π of two π, three π will equal πΎ times two π cubed over three π. We can distribute this cube and say two cubed times π cubed. So we can say that π of two π, three π equals πΎ times two cubed times π cubed times three π.

The πΎ stays the same. Two cubed equals eight. π cubed equals π cubed. And then in our denominator, we have three times π. Can we regroup πΎ times eight times π cubed over three times π to use the information from π of π, π? If we take out the eight-thirds, we could say that π of two π, three π is equal to eight-thirds times πΎπ cubed over π, which is π of π, π. And we know that that equals three. That means π of two π, three π is equal to eight-thirds times three. The three in the numerator and the three in the denominator cancel out, leaving us with eight.

Based on the given information, we can say that π of two π, three π equals eight.